A standard sheet of paper in Europe is 21.0 cm wide and 29.7 cm long. Which has the greater area, the standard sheet of paper in
1 answer:
You might be interested in
Answer:
84 feet
Step-by-step explanation:
This problem involves several steps. The first step is to realize that the given figure does not show the required number of vertical stringers. It shows 5, but there will be 7 of them. The given diagram is helpful in that it shows a vertical stringer on the centerline of the arches.
The second step is to write a function that will tell you how long the stringer will be. I find it convenient to write the equation for an arch shape such as this using the parent function h(x) = 1-x^2. This parent function gives an arch of height 1 and width 1 from center (a total width of 2). You want an arch that is 16 ft high and 40 ft wide (one side from center), so you must scale this parent function both horizontally (by 40) and vertically (by 16). It becomes ...
H(x) = 16(1 -(x/40)^2)
The taller arch is twice this height, so the length of a vertical stringer at position x is
vertical length = 2H(x) -H(x) = H(x)
That is, the function H(x) we defined can be used to find the length of the stringers.
The third step is to find the stringer lengths. It can save some energy if you realize that the problem is symmetrical, so that the stringer at x=-30 is the same length as the one at x=30. We need to find stringer lengths every 10 feet from -40 feet to +40 feet. Of course, the ones at ±40 feet are zero length, because that is where the two arches meet.
H(-30) = 16(1 -(3/4)^2) = 7
H(-20) = 16(1 -(1/2)^2) = 12
H(-10) = 16(1 -(1/4)^2) = 15
H(0) = 16
Then the fourth step is to add up the stringer lengths, rounding the result as required.
7 +12 +15 +16 +15 +12 +7 = 16 + 2(34) = 84 . . . . feet (no rounding needed)
Finally, you need to answer the question asked:
The sum of vertical stringer lengths is 84 feet.
